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guapka [62]
3 years ago
13

Find the equation of the line passing through the points (6,2)(10,6)

Mathematics
2 answers:
Butoxors [25]3 years ago
8 0

Slope-intercept form: y= mx + b (m is the slope, b is the y-intercept or the y value when x = 0 ---> (0, y))

To find the slope, use the slope formula and plug in 2 points:

m=\frac{y_2-y_1}{x_2-x_1}

(x₁ , y₁) = (5, 2)

(x₂ , y₂) = (10, 6)

m=\frac{6-2}{10-5} =\frac{4}{5}

y=\frac{4}{5}x+b    To find b, plug in a point into the equation (5, 2)

2=\frac{4}{5}(5)+b

2 = 4 + b

-2 = b

y=\frac{4}{5}x -2

kramer3 years ago
6 0

Answer:

y = 4/5x -2

Step-by-step explanation:

equation of a line passing through two points is given by

y - y₁ = m (x - x₁), where m = (y₂ - y₁) / (x₂ - x₁)

y₂ = 6, y₁ = 2

x₂ = 10, x₁ =5

m = (6-2)/(10-5)

m = 4/5

y - 2 = 4/5 (x - 5)

multiply both sides by 5

5(y -2) = 4(x - 5)

5y -10 = 4x -20

5y = 4x -20 +10

5y = 4x -10

divide through by 5

y =4/5x -2

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Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean
LuckyWell [14K]

Answer:

t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240

df = n_1 +n_2 -2 = 10+15-2= 23

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Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

Step-by-step explanation:

Data given

\bar X_1 = 1085 sample mean for group 1

\bar X_2 = 1034 sample mean for group 2

n_1 = 10 sample size for group 1

n_2 = 15 sample size for group 2

s_1 = 52 sample deviation for group 1

s_2 = 61 sample deviation for group 2

Solution

We want to check if the two means are equal so then the system of hypothesis are:

Null hypothesis: \mu_1= \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

And the statistic is given by:

t = \frac{\bar X_1 -\bar X_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}

And replacing we got:

t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240

The degrees of freedom are given by:

df = n_1 +n_2 -2 = 10+15-2= 23

And the p value would be:

p_v = 2*P(t_{23} >2.240) = 0.035

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

8 0
4 years ago
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